$\alpha$, $\beta$, $\gamma$ are the root of $x^3-x^2+px-1=0$. $(\alpha^3+1)(\beta^3+1)(\gamma^3+1)=2019$. $\alpha$, $\beta$, $\gamma$ are the root of $x^3-x^2+px-1=0$. $(\alpha^3+1)(\beta^3+1)(\gamma^3+1)=2019$. What is the product of all possible value of $p$?
Note that $p$ could be a complex number. 
I tried some basic Vieta's formulas, couldn't find easy way to simplify... 
 A: $$x^3-x^2+px-1=0 \Rightarrow (x^3-1)^3-(x^2-px)^3=0 \Rightarrow x^9+(3p-.4)x^6+(p^3-3p+3)-1=0.$$ 
Let us transform this equation by $y=x^3+1 \rightarrow x=(y-1)^{1/3}$. Then
we get a cubic Eq. for $y$ as
$$y^3+y^2(3p-7)+(p^3-9p+14)y-p^3+6p-9=0,$$ $y_1, y_2, y_3$ are its roots.
Then $$y_1 y_2 y_3=p^3-6p+9=2019 \Rightarrow p^3-6p-2010=0.$$
The roots of $p$ are $p_1,p_2,p_3$ and their product is: $p_1p_2p_3=2010$
A: $$(x^3-1)^3=(x^2-px)^3$$
$$(x^3)^3-1-3x^3(x^3-1)=(x^3)^2-p^3(x^3)-3px^3(x^3-1)$$
Replace $x^3+1$ with $y$
$$(y-1)^3-3(y-1)(y-2)=(y-1)^2-p^3(y-1)-3p(y-1)(y-2)$$
$$y^3+(\cdots)y^2+(\cdots)y-1-6+1-p^3+6p=0$$
Now apply Vieta's formula
A: $\DeclareMathOperator{\Res}{Res}$
Lazy answer, in line with some of the comments: let
$$\begin{align}
f(x)&=x^3-x^2+px-1\\
g(x)&=x^3+1\text{.}
\end{align}$$
Then the given is
$$\Res(f,g)=2019\text{,}$$
where $\Res$ denotes the resultant of $f$ and $g$.
Resultants are implemented in most computer algebra systems. In this case, we find
$$\Res(f,g)=p^3-6p+9$$
so
$$p^3-6p-2010=0\text{.}$$
Since the discriminant of the polynomial on the left is non-zero, the product of possible values for $p$ is equal to $2010$.
